Problem Statement
A trapezoidal channel with bottom slope 0.000169, bottom width 10 m and side slopes 1 : 1 carries 20 m³/s when Manning's constant = 0.015. Determine the normal depth.
Given Data & Constants
- Discharge, \(Q = 20 \, \text{m}^3/\text{s}\)
- Bed slope, \(i = 0.000169\)
- Bed width, \(B = 10 \, \text{m}\)
- Side slope = 1 Horizontal to 1 Vertical, so \(n = 1\)
- Manning's roughness coefficient, \(N = 0.015\)
Solution
1. Set up the Manning's Equation
The discharge in a channel is given by Manning's formula. We need to find the depth, \(d\), that satisfies this equation for the given discharge.
$$ Q = A \times \frac{1}{N} m^{2/3} i^{1/2} $$
$$ \text{Where:} $$
$$ \text{Area, } A = (B + nd)d = (10 + d)d $$
$$ \text{Wetted Perimeter, } P = B + 2d\sqrt{1+n^2} = 10 + 2d\sqrt{1+1^2} = 10 + 2\sqrt{2}d $$
$$ \text{Hydraulic Mean Depth, } m = \frac{A}{P} = \frac{(10+d)d}{10 + 2\sqrt{2}d} $$
Substituting the known values into the main equation gives:
$$ 20 = ((10+d)d) \times \frac{1}{0.015} \times \left(\frac{(10+d)d}{10 + 2.828d}\right)^{2/3} \times (0.000169)^{1/2} $$
$$ 20 = ((10+d)d) \times 66.67 \times \left(\frac{(10+d)d}{10 + 2.828d}\right)^{2/3} \times 0.013 $$
$$ \frac{20}{66.67 \times 0.013} = \frac{((10+d)d)^{5/3}}{(10 + 2.828d)^{2/3}} $$
$$ 23.07 \approx \frac{(10d+d^2)^{5/3}}{(10 + 2.828d)^{2/3}} $$
2. Solve by Trial and Error
We must now guess values for \(d\) and see which one satisfies the equation. Let's test a few values.
Try d = 1.5 m:
$$ \text{Right Hand Side} = \frac{(10(1.5)+1.5^2)^{5/3}}{(10 + 2.828(1.5))^{2/3}} = \frac{(17.25)^{5/3}}{(14.242)^{2/3}} \approx 16.9 $$
This is too low. We need to increase the depth.
Try d = 1.7 m:
$$ \text{Right Hand Side} = \frac{(10(1.7)+1.7^2)^{5/3}}{(10 + 2.828(1.7))^{2/3}} = \frac{(19.89)^{5/3}}{(14.808)^{2/3}} \approx 20.9 $$
This is closer. The correct depth is between 1.5 m and 1.7 m.
Try d = 1.65 m:
$$ \text{Right Hand Side} = \frac{(10(1.65)+1.65^2)^{5/3}}{(10 + 2.828(1.65))^{2/3}} = \frac{(19.22)^{5/3}}{(14.666)^{2/3}} \approx 19.9 $$
This is very close to the required value of 20.
Final Result:
The normal depth of the channel is approximately 1.65 m.
